3
Vectors
CHAPTER OUTLINE
3.1
Coordinate Systems
3.2
Vector and Scalar Quantities
3.3
Some Properties of Vectors
3.4
Components of a Vector and Unit Vectors
* An asterisk indicates a question or problem new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
102 + 102 m/s.
OQ3.1
Answer (e). The magnitude is
OQ3.2
Answer (e). If the quantities x and y are positive, a vector with
components (−x, y) or (x, −y) would lie in the second or fourth
quadrant, respectively.


Answer (a). The vector −2D1 will be twice as long as D1 and in the

opposite direction, namely northeast. Adding D2 , which is about
equally long and southwest, we get a sum that is still longer and due
east.
*OQ3.3
OQ3.4
The ranking is c = e > a > d > b. The magnitudes of the vectors being
added are constant, and we are considering the magnitude only—not
the direction—of the resultant. So we need look only at the angle
between the vectors being added in each case. The smaller this angle,
the larger the resultant magnitude.
OQ3.5
Answers (a), (b), and (c). The magnitude can range from the sum of the
individual magnitudes, 8 + 6 = 14, to the difference of the individual
magnitudes, 8 − 6 = 2. Because magnitude is the “length” of a vector, it
is always positive.
98
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Chapter 3
OQ3.6
99

Answer (d). If we write vector A as
( A , A ) = ( − A
x
y
x
, Ay
)
and vector B as
(B , B ) = ( B
x
y
x
, − By
)
then
(
) (
 
B − A = Bx − ( − Ax ) , − By − Ay = Bx + Ax , − By − Ay
which would be in the fourth quadrant.
)
OQ3.7
The answers are (a) yes (b) no (c) no (d) no (e) no (f) yes (g) no. Only
force and velocity are vectors. None of the other quantities requires a
direction to be described.
OQ3.8
Answer (c). The vector has no y component given. It is therefore 0.
OQ3.9
Answer (d). Take the difference of the x coordinates of the ends of the
vector, head minus tail: –4 – 2 = –6 cm.
OQ3.10
Answer (a). Take the difference of the y coordinates of the ends of the
vector, head minus tail: 1 − (−2) = 3 cm.
OQ3.11
Answer (c). The signs of the components of a vector are the same as the
signs of the points in the quadrant into which it points. If a vector
arrow is drawn to scale, the coordinates of the point of the arrow equal
the components of the vector. All x and y values in the third quadrant
are negative.
OQ3.12
Answer (c). The vertical component is opposite the 30° angle, so
sin 30° = (vertical component)/50 m.
OQ3.13
Answer (c). A vector in the second quadrant has a negative x
component and a positive y component.
ANSWERS TO CONCEPTUAL QUESTIONS
CQ3.1
Addition of a vector to a scalar is not defined. Try adding the speed
and velocity, 8.0 m/s + (15.0 m/s î) : Should you consider the sum to
be a vector or a scalar? What meaning would it have?
CQ3.2
No, the magnitude of a vector is always positive. A minus sign in a
vector only indicates direction, not magnitude.
CQ3.3
(a) The book’s displacement is zero, as it ends up at the point from
which it started. (b) The distance traveled is 6.0 meters.
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100
CQ3.4
CQ3.5
Vectors


Vectors A and B are perpendicular to each other.
The inverse tangent function gives the correct angle, relative to the +x
axis, for vectors in the first or fourth quadrant, and it gives an incorrect
answer for vectors in the second or third quadrant. If the x and y
components are both positive, their ratio y/x is positive and the vector
lies in the first quadrant; if the x component is positive and the y
component negative, their ratio y/x is negative and the vector lies in
the fourth quadrant. If the x and y components are both negative, their
ratio y/x is positive but the vector lies in the third quadrant; if the x
component is negative and the y component positive, their ratio y/x is
negative but the vector lies in the second quadrant.
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 3.1
P3.1
Coordinate Systems
ANS. FIG. P3.1 helps to visualize the x and y
coordinates, and trigonometric functions will
tell us the coordinates directly. When the polar
coordinates (r, θ) of a point P are known, the
Cartesian coordinates are found as
x = r cos θ and y = r sin θ
Then,
x = r cosθ = ( 5.50 m ) cos 240°
= ( 5.50 m )( −0.5 ) = −2.75 m
y = r sin θ = ( 5.50 m ) sin 240°
= ( 5.50 m )( −0.866 ) = −4.76 m
P3.2
(a)
We use x = r cosθ . Substituting, we have 2.00 = r cos 30.0°, so
r=
*P3.3
2.00
= 2.31
cos 30.0°
(b)
From y = r sin θ , we have y = r sin 30.0° = 2.31sin 30.0° = 1.15 .
(a)
The distance between the points is given by
d=
( x2 − x1 )2 + ( y2 − y1 )2
= ( 2.00 − [ −3.00 ])2 + ( −4.00 − 3.00 )2
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Chapter 3
101
d = 25.0 + 49.0 = 8.60 m
(b)
To find the polar coordinates of each point, we measure the radial
distance to that point and the angle it makes with the +x axis:
r1 = ( 2.00 )2 + ( −4.00 )2 = 20.0 = 4.47 m
(
θ 1 = tan −1 −
)
4.00
= −63.4°
2.00
r2 = ( −3.00 )2 + ( 3.00 )2 = 18.0 = 4.24 m
θ 2 = 135° measured from the +x axis.
P3.4
(a)
x = r cosθ and y = r sin θ , therefore,
x1 = (2.50 m) cos 30.0°, y1 = (2.50 m) sin 30.0°, and
(x1 , y1 ) = (2.17, 1.25) m
x2 = (3.80 m) cos 120°, y2 = (3.80 m) sin 120°, and
(x2 , y 2 ) = (−1.90, 3.29) m
(b)
P3.5
P3.6
d = (Δ x)2 + (Δ y)2 = 4.07 2 + 2.042 m = 4.55 m
For polar coordinates (r, θ), the Cartesian coordinates are (x = r cosθ,
y = r sinθ), if the angle is measured relative to the +x axis.
(a)
(–3.56 cm, – 2.40 cm)
(b)
(+3.56 cm, – 2.40 cm) → (4.30 cm, – 34.0°)
(c)
(7.12 cm, 4.80 cm) → (8.60 cm, 34.0°)
(d)
(–10.7 cm, 7.21 cm) → (12.9 cm, 146°)
⎛ y⎞
We have r = x 2 + y 2 and θ = tan −1 ⎜ ⎟ .
⎝ x⎠
(a)
The radius for this new point is
(−x)2 + y 2 = x 2 + y 2 = r
and its angle is
⎛ y ⎞
tan −1 ⎜ ⎟ = 180° − θ
⎝ −x ⎠
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102
Vectors
(b)
(−2x)2 + (−2y)2 = 2r . This point is in the third quadrant if (x, y)
is in the first quadrant or in the fourth quadrant if (x, y) is in the
second quadrant. It is at an angle of 180° + θ .
(c)
(3x)2 + (−3y)2 = 3r . This point is in the fourth quadrant if (x, y)
is in the first quadrant or in the third quadrant if (x, y) is in the
second quadrant. It is at an angle of −θ or 360 − θ .
Section 3.2
Vector and Scalar Quantities
Section 3.3
Some Properties of Vectors
P3.7
Figure P3.7 suggests a right triangle where, relative to angle θ, its
adjacent side has length d and its opposite side is equal to width of the
river, y; thus,
tan θ =
y
→ y = d tan θ
d
y = (100 m)tan(35.0°) = 70.0 m
P3.8
The width of the river is 70.0 m .
  
We are given R = A + B . When two vectors are
added graphically, the second vector is positioned
with its tail at the tip of the first vector. The resultant
then runs from the tail of the first vector to the tip of

the second vector. In this case, vector A will be
positioned with its tail at the origin and its tip at the
point (0, 29). The resultant is then drawn, starting at
the origin (tail of first vector) and going 14 units in
the negative y direction to the point (0, −14). The


second vector, B , must then start from the tip of A

at point (0, 29) and end on the tip of R at point
(0, −14) as shown in the sketch at the right. From
this, it is seen that

B is 43 units in the negative y direction
ANS. FIG. P3.8
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Chapter 3
P3.9
103
In solving this problem we must contrast
displacement with distance traveled. We draw a
diagram of the skater’s path in ANS. FIG. P3.9,
which is the view from a hovering helicopter so
that we can see the circular path as circular in
shape. To start with a concrete example, we have
chosen to draw motion ABC around one half of a ANS. FIG. P3.9
circle of radius 5 m.

The displacement, shown as d in the diagram, is the straight-line
change in position from starting point A to finish C. In the specific case
we have chosen to draw, it lies along a diameter of the circle. Its

magnitude is d = –10.0 î = 10.0 m.
The distance skated is greater than the straight-line displacement. The
distance follows the curved path of the semicircle (ABC). Its length is
1
half of the circumference: s = (2π r) = 5.00π m = 15.7 m.
2
A straight line is the shortest distance between two points. For any
nonzero displacement, less or more than across a semicircle, the
distance along the path will be greater than the displacement
magnitude. Therefore:
The situation can never be true because the distance is
an arc of a circle between two points, whereas the
magnitude of the displacement vector is a straight-line
cord of the circle between the same points.
P3.10
 
We find the resultant F1 + F2 graphically by


placing the tail of F2 at the head of F1 . The
 
resultant force vector F1 + F2 is of magnitude
9.5 N and at an angle of
57° above the x axis .
ANS. FIG. P3.10
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104
P3.11
Vectors
To find these vector expressions
graphically, we draw each set of
vectors. Measurements of the
results are taken using a ruler
and protractor. (Scale: 1 unit =
0.5 m)
 
(a) A + B = 5.2 m at 60o
 
(b) A − B = 3.0 m at 330o
 
(c) B − A = 3.0 m at 150o

(d) A − 2B = 5.2 m at 300o
ANS. FIG. P3.11
P3.12
(a)
The three diagrams are shown in ANS. FIG. P3.12a below.
ANS. FIG. P3.12a
(b)
P3.13
The diagrams in ANS. FIG. P3.12a represent the graphical
   

  
solutions for the three vector sums: R 1 = A + B + C, R 2 = B + C + A,
  

and R 3 = C + B + A.
The scale drawing for the
graphical solution should be
similar to the figure to the
right. The magnitude and
direction of the final
displacement from the
starting point are obtained
ANS. FIG. P3.13
by measuring d and θ on the
drawing and applying the scale factor used in making the drawing.
The results should be d = 420 ft and θ = –3° .
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Chapter 3
*P3.14
105
ANS. FIG. P3.14 shows the graphical
addition of the vector from the base camp to
lake A to the vector connecting lakes A and
B, with a scale of 1 unit = 20 km. The
distance from lake B to base camp is then
the negative of this resultant vector, or

−R = 310 km at 57° S of W .
ANS. FIG. P3.14
Section 3.4
P3.15
Components of a Vector
and Unit Vectors
First we should visualize the vector either in our
mind or with a sketch, as shown in ANS. FIG.
P3.15. The magnitude of the vector can be found
by the Pythagorean theorem:
Ax = –25.0
Ay = 40.0
A = A + A = (−25.0) + (40.0)
2
x
2
y
2
2
ANS. FIG. P3.15
= 47.2 units
We observe that
tan φ =
Ay
Ax
so
⎛ Ay ⎞
⎛ 40.0 ⎞
φ = tan −1 ⎜
= tan −1 ⎜
= tan −1 (1.60) = 58.0°
⎟
⎟
⎝ 25.0 ⎠
⎝ Ax ⎠
The diagram shows that the angle from the +x axis can be found by
subtracting from 180°: θ = 180° − 58° = 122°
P3.16
We can calculate the components of the vector A using (Ax, Ay) =
(A cos θ, A sin θ) if the angle θ is measured from the +x axis, which is
true here. For A = 35.0 units and θ = 325°,
Ax = 28.7 units, Ay = –20.1 units
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106
P3.17
Vectors
(a) Yes .
(b)
Let v represent the speed of the camper. The northward
component of its velocity is v cos 8.50°. To avoid crowding the
minivan we require v cos 8.50° ≥ 28 m/s.
We can satisfy this requirement simply by taking
v ≥ (28.0 m/s)/cos 8.50° = 28.3 m/s.
P3.18
The person would have to walk
( 3.10 km ) sin 25.0° = 1.31 km north
and
P3.19
P3.20
( 3.10 km ) cos 25.0° = 2.81 km east
Do not think of sin θ = opposite/hypotenuse, but jump right to y =
R sin θ. The angle does not need to fit inside a triangle. We find the x
and y components of each vector using x = r cos θ and y = r sin θ. In

unit vector notation, R = Rx î + Ry ĵ.
(a)
x = 12.8 cos 150°, y = 12.8 sin 150°, and (x, y) = (−11.1î + 6.40 ĵ) m
(b)
x = 3.30 cos 60.0°, y = 3.30 sin 60.0°, and (x, y) = (1.65î + 2.86 ĵ) cm
(c)
x = 22.0 cos 215°, y = 22.0 sin 215°, and (x, y) = (−18.0î − 12.6 ĵ) in
(a)
Her net x (east-west) displacement is –3.00 + 0 + 6.00 = +3.00
blocks, while her net y (north-south) displacement is 0 + 4.00 + 0 =
+4.00 blocks. The magnitude of the resultant displacement is
R = (xnet )2 + (y net )2 = (3.00)2 + (4.00)2 = 5.00 blocks
and the angle the resultant makes with the x axis (eastward
direction) is
⎛ 4.00 ⎞
θ = tan −1 ⎜
= tan −1 (1.33) = 53.1°.
⎝ 3.00 ⎟⎠
The resultant displacement is then 5.00 blocks at 53.1° N of E .
(b)
P3.21
The total distance traveled is 3.00 + 4.00 + 6.00 = 13.00 blocks .
Let +x be East and +y be North. We can sum the total x and y
displacements of the spelunker as
∑ x = 250 m + ( 125 m ) cos 30° = 358 m
∑ y = 75 m + ( 125 m ) sin 30° − 150 m = −12.5 m
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Chapter 3
107
the total displacement is then
d=
( ∑ x )2 + ( ∑ y )2 =
(358 m)2 + (−12.5 m)2 = 358 m
at an angle of
⎛ ∑y⎞
⎛ 12.5 m ⎞
θ = tan −1 ⎜
= tan −1 ⎜ −
= −2.00°
⎟
⎝ 358 m ⎟⎠
⎝ ∑x⎠

d = 358 m at 2.00° S of E
or
P3.22
We use the numbers given in Problem 3.11:

A = 3.00 m, θ A = 30.0°
Ax = A cos θA = 3.00 cos 30.0° = 2.60 m,
Ay = A sin θA = 3.00 sin 30.0° = 1.50 m

A = Ax î + Ay ĵ = (2.60î + 1.50 ĵ) m

B = 3.00 m, θ B = 90.0°

Bx = 0, By = 3.00 m → B = 3.00 ĵ m
So
 
A + B = 2.60î + 1.50 ĵ + 3.00 ĵ = 2.60î + 4.50 ĵ m
(
then
P3.23
)
(
)
We can get answers in unit-vector form just by doing calculations with
each term labeled with an î or a ĵ. There are, in a sense, only two
vectors to calculate, since parts (c), (d), and (e) just ask about the
magnitudes and directions of the answers to (a) and (b). Note that
the whole numbers appearing in the problem statement are assumed
to have three significant figures.
We use the property of vector addition that states that the components
  
of R = A + B are computed as Rx = Ax + Bx and Ry = Ay + By .




(a)
( A + B) = ( 3î − 2 ĵ) + ( − î − 4 ĵ) = 2î − 6 ĵ
(b)
( A − B) = ( 3î − 2 ĵ) − ( − î − 4 ĵ) = 4î + 2 ĵ
(c)
 
A + B = 2 2 + 62 = 6.32
(d)
 
A − B = 42 + 2 2 = 4.47
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108
Vectors
(e)
⎛ 6⎞
θ A +B = tan −1 ⎜ − ⎟ = −71.6° = 288°
⎝ 2⎠
⎛ 2⎞
θ A −B = tan −1 ⎜ ⎟ = 26.6°
⎝ 4⎠
P3.24
The east and north components of the displacement from Dallas (D) to
Chicago (C) are the sums of the east and north components of the
displacements from Dallas to Atlanta (A) and from Atlanta to Chicago.
In equation form:
dDC east = dDA east + dAC east
= ( 730 mi ) cos 5.00° − ( 560 mi ) sin 21.0° = 527 miles
dDC north = dDA north + dAC north
= ( 730 mi ) sin 5.00° + ( 560 mi ) cos 21.0° = 586 miles
By the Pythagorean theorem,
d = (dDC east )2 + (dDC north )2 = 788 mi
Then,
⎛d
⎞
θ = tan −1 ⎜ DC north ⎟ = 48.0°
⎝ dDC east ⎠
Thus, Chicago is 788 miles at 48.0° northeast of Dallas .
P3.25
We use the unit-vector addition method. It is just as easy to add three
displacements as to add two. We take the direction east to be along + î.
The three displacements can be written as:

d1 = ( −3.50 m ) ĵ

d2 = ( 8.20 m ) cos 45.0°î + ( 8.20 m ) sin 45.0° ĵ
= ( 5.80 m ) î + ( 5.80 m ) ĵ

d3 = ( −15.0 m ) î
The resultant is
   
R = d1 + d2 + d3 = (−15.0 m + 5.80 m)î + (5.80 m − 3.50 m) ĵ
= ( −9.20 m ) î + ( 2.30 m ) ĵ
(or 9.20 m west and 2.30 m north).
The magnitude of the resultant displacement is

2
2
R = Rx2 + Ry2 = ( −9.20 m ) + ( 2.30 m ) = 9.48 m
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Chapter 3
109
The direction of the resultant vector is given by
⎛ Ry ⎞
⎛ 2.30 m ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
= 166°
⎝ −9.20 m ⎟⎠
⎝ Rx ⎠
P3.26
(a)
(b)
See figure to the right.   
C = A + B = 2.00î + 6.00 ĵ + 3.00î − 2.00 ĵ = 5.00î + 4.00 ĵ
  
D = A − B = 2.00î + 6.00 ĵ − 3.00î + 2.00 ĵ
= −1.00î + 8.00 ĵ
(c)
P3.27

⎛ 4⎞
C = 25.0 + 16.0 at tan −1 ⎜ ⎟ = 6.40 at 38.7° ⎝ 5⎠

2
2
⎛ 8.00 ⎞
D = ( −1.00 ) + ( 8.00 ) at tan −1 ⎜
⎝ −1.00 ⎟⎠

D = 8.06 at ( 180° − 82.9° ) = 8.06 at 97.2°
ANS. FIG. P3.26
We first tabulate the three strokes of the novice golfer, with the x
direction corresponding to East and the y direction corresponding to
North. The sum of the displacement in each of the directions is shown
as the last row of the table.
East
North
x (m)
y (m)
0
4.00
1.41
1.41
−0.500
−0.866
+0.914
4.55
The “hole-in-one” single displacement is then

2
2
2
2
R = x + y = ( 0.914 m ) + ( 4.55 m ) = 4.64 m
The angle of the displacement with the horizontal is
⎛ y⎞
⎛ 4.55 m ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
= 78.6°
⎝ x⎠
⎝ 0.914 m ⎟⎠
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110
Vectors
The expert golfer would accomplish the hole in one with the
displacement 4.64 m at 78.6° N of E .
P3.28
We take the x axis along the slope downhill. (Students, get used to this
choice!) The y axis is perpendicular to the slope, at 35.0° to the vertical.
Then the displacement of the snow makes an angle of 90.0° + 35.0° +
16.0° = 141° with the x axis.
ANS. FIG. P3.28
(a)
Its component parallel to the surface is
(1.50 m) cos141° = –1.17 m,
or 1.17 m toward the top of the hill.
(b)
Its component perpendicular to the surface is
(1.50 m)sin 141° = 0.944 m, or 0.944 m away
from the snow.
P3.29
(a)
The single force is obtained by summing the two forces:
  
F = F1 + F2

F = 120 cos (60.0°)î + 120 sin (60.0°) ĵ
− 80.0 cos (75.0°)î + 80.0 sin (75.0°) ĵ

F = 60.0î + 104 ĵ − 20.7 î + 77.3 ĵ = 39.3î + 181ĵ N
(
)
We can also express this force in terms of its magnitude and
direction:

F = 39.32 + 1812 N = 185 N
⎛ 181 ⎞
θ = tan −1 ⎜
= 77.8°
⎝ 39.3 ⎟⎠
(b)
A force equal and opposite the resultant force from part (a) is
required for the total force to equal zero:


F3 = − F = −39.3î − 181ĵ N (
)
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.30
111
ANS. FIG. P3.30 is a graphical depiction of the three displacements

the football undergoes,
with
corresponding to the 10.0-yard
A

backward run, B corresponding to the 15.0-yard sideways run, and

C corresponding to the 50.0-yard downfield pass. The resultant
vector is then
   
R = A + B + C = −10.0î − 15.0 ĵ + 50.0î
= 40.0î − 15.0 ĵ

1/2
R = ⎡⎣(40.0)2 + (−15.0)2 ⎤⎦ = 42.7 yards
ANS. FIG. P3.30
P3.31
P3.32
P3.33
(a)
We add the components of the three vectors:
   
D = A + B + C = 6î − 2 ĵ

D = 62 + 2 2 = 6.32 m at θ = 342°
(b)
Again, using the components of the three vectors,
  

E = − A − B + C = −2 î + 12 ĵ

E = 2 2 + 12 2 = 12.2 m at θ = 99.5°


We are given A = −8.70î + 15.0 ĵ, and B = 13.2 î − 6.60 ĵ, and
 


A − B + 3C = 0. Solving for C gives
  
3C = B − A = 21.9î − 21.6 ĵ

C = 7.30î − 7.20 ĵ or Cx = 7.30 cm ; Cy = −7.20 cm
Hold your fingertip at the center of the front edge of your study desk,
defined as point O. Move your finger 8 cm to the right, then 12 cm
vertically up, and then 4 cm horizontally away from you. Its location

relative to the starting point represents position vector A. Move threefourths of the way straight back toward O. Now your fingertip is at the

location of B. Now move your finger 50 cm straight through O,
through your left thigh, and down toward the floor. Its position vector

now is C.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
112
Vectors
We use unit-vector notation throughout. There is no adding to do here,
but just multiplication of a vector by two different scalars.

(a) A = 8.00î + 12.0 ĵ − 4.00k̂ 
 A
(b) B = = 2.00î + 3.00 ĵ − 1.00k̂ 4


(c) C = −3A = −24.0 î − 36.0 ĵ + 12.0k̂ P3.34

We are given B = Bx î + By ĵ + Bz k̂ = 4.00î + 6.00 ĵ + 3.00k̂. The magnitude
of the vector is therefore

B = 4.002 + 6.002 + 3.002 = 7.81
And the angle of the vector with the three coordinate axes is
⎛ 4.00 ⎞
α = cos −1 ⎜
= 59.2° is the angle with the x axis
⎝ 7.81 ⎟⎠
⎛ 6.00 ⎞
β = cos −1 ⎜
= 39.8° is the angle with the y axis
⎝ 7.81 ⎟⎠
P3.35
⎛ 3.00 ⎞
γ = cos −1 ⎜
= 67.4° is the angle with the z axis
⎝ 7.81 ⎟⎠

The component description of A is just restated to constitute the
answer to part (a): Ax= −3.00, Ay = 2.00.

(a) A = Ax î + Ay ĵ = −3.00î + 2.00 ĵ
(b)

A = Ax2 + Ay2 =
( −3.00)2 + ( 2.00)2
= 3.61
⎛ Ay ⎞
⎛ 2.00 ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
= −33.7°
⎝ −3.00 ⎟⎠
⎝ Ax ⎠
(c)
θ is in the second quadrant, so θ = 180° + ( −33.7° ) = 146° .
  
  
Rx = 0, Ry = −4.00, and R = A + B, thus B = R − A and
Bx = Rx – Ax = 0 – (–3.00) = 3.00, By = Ry – Ay= –4.00 – 2.00 = –6.00.

Therefore, B = 3.00î − 6.00 ĵ .
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.36
113
We carry out the prescribed mathematical operations using unit
vectors.
  
(a) C = A + B = 5.00î − 1.00 ĵ − 3.00k̂ m
(
)

2
2
2
C = ( 5.00 m ) + ( 1.00 m ) + ( 3.00 m ) = 5.92 m
(b)
 

D = 2 A − B = 4.00î − 11.0 ĵ + 15.0k̂ m
(
)

D = (4.00 m)2 + (11.0 m)2 + (15.0 m)2 = 19.0 m
P3.37
(a)
Taking components along î and ĵ, we get two equations: 6.00a – 8.00b +26.0 = 0
and –8.00a + 3.00b + 19.0 = 0 Substituting a = 1.33b – 4.33 into the second equation, we find −8 ( 1.33b − 4.33 ) + 3b + 19 = 0 → 7.67b = 53.67 → b = 7.00 and so a = 1.33(7.00) – 4.33 = 5.00. 
 
Thus a = 5.00, b = 7.00 . Therefore, 5.00A + 7.00B + C = 0. (b)
P3.38
In order for vectors to be equal, all of their components must be equal. A vector equation contains more information than a
scalar equation, as each component gives us one equation.
The given diagram shows the vectors individually,
but not their addition. The second diagram
represents a map view of the motion of the ball.
According to the definition of a displacement, we
ignore any departure from straightness of the actual
path of the ball. We model each of the three motions
as straight. The simplified problem is solved by
straightforward application of the component
method of vector addition. It works for
adding two, three, or any number of vectors.
(a)
We find the two components of each of
the three vectors
Ax = (20.0 units)cos 90° = 0
and Ay = (20.0 units)sin 90° = 20.0 units
ANS. FIG. P3.38
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
114
Vectors
Bx = (40.0units)cos 45° = 28.3 units
and By = (40.0units)sin 45° = 28.3 units
Cx = (30.0units)cos 315° = 21.2 units
and Cy = (30.0units)sin 315° = −21.2 units
Now adding,
Rx = Ax + Bx + Cx = (0 + 28.3 + 21.2) units = 49.5 units
and Ry = Ay + By + Cy = (20 + 28.3 − 21.2) units = 27.1 units
so
(b)

R = 49.5î + 27.1ĵ 
2
2
R = ( 49.5 ) + ( 27.1) = 56.4
⎛ Ry ⎞
⎛ 27.1 ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
= 28.7°
⎝ 49.5 ⎟⎠
⎝ Rx ⎠
P3.39
We will use the component method for a precise
answer. We already know the total displacement,
so the algebra of solving a vector equation will
guide us to do a subtraction.
  
We have B = R − A: Ax = 150 cos 120° = −75.0 cm
ANS. FIG. P3.39
Ay = 150 sin 120° = 130 cm
Rx = 140 cos 35.0° = 115 cm
Ry = 140 sin 35.0° = 80.3 cm
Therefore, 
B = [115 − (−75)î + [80.3 − 130]ĵ = 190î − 49.7 ĵ cm

B = 1902 + 49.7 2 = 196 cm
(
)
⎛ 49.7 ⎞
θ = tan −1 ⎜ −
= −14.7°
⎝ 190 ⎟⎠
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.40
115
First, we sum the components of the two vectors for the male:
d 3mx = d1mx + d 2mx = 0 + ( 100 cm ) cos 23.0° = 92.1 cm
d 3my = d1my + d 2my = 104 cm + ( 100 cm ) sin 23.0° = 143.1 cm
magnitude: d 3m = ( 92.1 cm ) + ( 143.1 cm ) = 170.1 cm
2
2
direction: tan −1 ( 143.1/ 92.1) = 57.2° above +x axis ( first quadrant )
followed by the components of the two vectors for the female:
d 3fx = d1fx + d 2fx = 0 + ( 86.0 cm ) cos 28.0° = 75.9 cm
d 3fy = d1fy + d 2fy = 84.0 cm + ( 86.0 cm ) sin 28.0° = 124.4 cm
magnitude: d 3f = ( 75.9 cm ) + ( 124.4 cm ) = 145.7 cm
2
2
direction: tan −1 ( 124.4/75.9 ) = 58.6° above +x axis ( first quadrant )
P3.41
(a)

E = (17.0 cm) cos ( 27.0° ) î
+ (17.0 cm) sin ( 27.0° ) ĵ

E = (15.1î + 7.72 ĵ) cm
(b)

F = (17.0 cm) cos ( 117.0° ) î
+ (17.0 cm) sin ( 117.0° ) ĵ

F = −7.72 î + 15.1ĵ cm
(
(c)
)
Note that we did not need to explicitly identify the angle with the
positive x axis, but by doing so, we don’t have to keep track of
minus signs for the components. 
G = [(−17.0 cm) cos ( 243.0° )] î + [(−17.0 cm) sin ( 243.0° )] ĵ

G = −7.72 î − 15.1ĵ cm
(
P3.42
ANS. FIG. P3.41
)
The position vector from radar station to ship is

S = 17.3sin 136°î + 17.3cos136° ĵ km = 12.0î − 12.4 ĵ km
(
)
(
)
From station to plane, the position vector is

P = 19.6sin 153°î + 19.6cos153° ĵ + 2.20k̂ km
(
)
or
(
)

P = 8.90î − 17.5 ĵ + 2.20k̂ km
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
116
Vectors
(a)
To fly to the ship, the plane must undergo displacement
  
D = S − P = 3.12 î + 5.02 ĵ − 2.20k̂ km
(
(b)
P3.43
)
The distance the plane must travel is

2
2
2
D = D = ( 3.12 ) + ( 5.02 ) + ( 2.20 ) km = 6.31 km
The hurricane’s first displacement is
( 41.0 km/h )( 3.00 h ) at 60.0° N of W
and its second displacement is
( 25.0 km/h )(1.50 h ) due North
With î representing east and ĵ representing north, its total
displacement is:
[( 41.0 km/h ) cos60.0°]( 3.00 h )( − î )
+ [( 41.0 km/h ) sin 60.0° ]( 3.00 h ) ĵ
+ ( 25.0 km/h ) ( 1.50 h ) ĵ
( )
= 61.5 km − î + 144 km ĵ
with magnitude
P3.44
(61.5 km )2 + (144 km )2
= 157 km .
Note that each shopper must make a choice whether to turn 90° to the
left or right, each time he or she makes a turn. One set of such choices,
following the rules in the problem, results in the shopper heading in
the positive y direction and then again in the positive x direction.
Find the magnitude of the sum of the displacements:

d = (8.00 m)î + (3.00 m) ĵ + (4.00 m)î = (12.00 m)î + (3.00 m) ĵ
magnitude: d = (12.00 m)2 + (3.00 m)2 = 12.4 m
Impossible because 12.4 m is greater than 5.00 m.
P3.45
The y coordinate of the airplane is constant and equal to 7.60 × 103 m
whereas the x coordinate is given by x = vit, where vi is the constant
speed in the horizontal direction.
At t = 30.0 s we have x = 8.04 × 103, so vi = 8 040 m/30 s = 268 m/s. The
position vector as a function of time is

P = (268 m/s)tî + (7.60 × 103 m) ĵ
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
117

At t = 45.0 s, P = ⎡⎣1.21 × 10 4 î + 7.60 × 103 ĵ ⎤⎦ m. The magnitude is

P=
(1.21 × 10 ) + (7.60 × 10 )
4 2
3 2
m = 1.43 × 10 4 m
and the direction is
⎛ 7.60 × 103 ⎞
θ = tan −1 ⎜
= 32.2° above the horizontal
⎝ 1.21 × 10 4 ⎟⎠
P3.46
The displacement from the start to the finish is
16î + 12 ĵ − (5î + 3 ĵ) = (11î + 9 ĵ)
(
)
The displacement from the starting point to A is f 11î + 9 ĵ meters.
(a)
The position vector of point A is
(
)
5î + 3 ĵ + f 11î + 9 ĵ = ⎡⎣(5 + 11 f )î + (3 + 9 f ) ĵ ⎤⎦ m
(b)
(c)
For f = 0 we have the position vector (5 + 0)î + (3 + 0) ĵ meters.
This is reasonable because it is the location of the starting point,
5î + 3 ĵ meters.
(d) For f = 1 = 100%, we have position vector
(5 + 11)î + (3 + 9) ĵ meters = 16î + 12 ĵ meters .
(e)
P3.47
This is reasonable because we have completed the trip, and this is
the position vector of the endpoint.
Let the positive x direction be eastward, the positive y direction be
vertically upward, and the positive z direction be southward. The total
displacement is then 
d = 4.80î + 4.80 ĵ cm + 3.70 ĵ − 3.70k̂ cm
(
) (
= ( 4.80î + 8.50 ĵ − 3.70k̂ ) cm
)
( 4.80)2 + ( 8.50)2 + ( −3.70)2
(a)
The magnitude is d =
(b)
Its angle with the y axis follows from
8.50
cosθ =
, giving θ = 35.5° .
10.4
cm = 10.4 cm . © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
118
Vectors
Additional Problems
P3.48
The Pythagorean theorem and the definition of the tangent will be the
starting points for our calculation.
(a)
Take the wall as the xy plane so that the coordinates are x = 2.00 m
and y = 1.00 m; and the fly is located at point P. The distance
between two points in the xy plane is
d=
(x
so here d =
(b)
P3.49
– x1 ) + ( y 2 – y1 )
2
2
2
(2.00 m – 0)2 + (1.00 m – 0)2 = 2.24 m
⎛y⎞
⎛ 1.00 m ⎞⎟

°
θ = tan –1 ⎜⎜ ⎟⎟⎟ = tan –1 ⎜⎜
=
26.6
,
so
r
= 2.24 m, 26.6°
⎟
⎝x⎠
⎝ 2.00 m ⎟⎠
We note that − î = west and − ĵ = south. The given
mathematical representation of the trip can be
written as 6.30 b west + 4.00 b at 40° south of west
+3.00 b at 50° south of east +5.00 b south. (a)
(b)
The figure on the right shows a map of the
successive displacements that the bus
undergoes. ANS. FIG. P3.49
The total odometer distance is the sum of the magnitudes of the
four displacements: 6.30 b + 4.00 b + 3.00 b + 5.00 b = 18.3 b (c)

R = (−6.30 − 3.06 + 1.93) b î + (−2.57 − 2.30 − 5.00) b ĵ
= −7.44 b î − 9.87 b ĵ
⎛ 9.87 ⎞
= (7.44 b)2 + (9.87 b)2 at tan −1 ⎜
south of west
⎝ 7.44 ⎟⎠
= 12.4 b at 53.0° south of west
= 12.4 b at 233° counterclockwise from east
P3.50
To find the new speed and direction of the aircraft, we add the
vector components of the wind to the vector velocity of the aircraft:

v = vx î + vy ĵ = ( 300 + 100cos 30.0° ) î + ( 100sin 30.0° ) ĵ

v = 387 î + 50.0 ĵ mi/h

v = 390 mi/h at 7.37° N of E
(
)
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.51
119
On our version of the diagram we have
drawn in the resultant from the tail of the
first arrow to the head of the last arrow.

The resultant displacement R is equal to
the sum of the four individual
 
  
displacements, R = d1 + d2 + d3 + d4 . We
translate from the pictorial representation
to a mathematical representation by
writing the individual displacements in
unit-vector notation:
ANS. FIG. P3.51


d1 = 100i m

d2 = −300j m

d3 = (−150 cos 30°)î m + ( − 150 sin 30°)ĵ m = –130î m − 75 ĵ m

d4 = (− 200 cos60°)î m + (200 sin 60°)ĵ m = − 100î m + 173 ĵ m
Summing the components together, we find
Rx = d1x + d2 x + d3x + d4x = (100 + 0 − 130 − 100) m = − 130 m
Ry = d1y + d2 y + d3y + d4y = (0 − 300 − 75 + 173) m = –202 m
so altogether
    
R = d1 + d2 + d3 + d4 = −130î − 202 ĵ m
(
)
Its magnitude is

2
2
R = ( −130 ) + ( −202 ) = 240 m
⎛R ⎞
−202 ⎞
We calculate the angle φ = tan −1 ⎜ y ⎟ = tan −1 ⎛⎜
= 57.2°.
⎝ −130 ⎟⎠
⎝ Rx ⎠
The resultant points into the third quadrant instead of the first
quadrant. The angle counterclockwise from the +x axis is
θ = 180 + φ = 237°
*P3.52
The superhero follows a straight-line path at
30.0° below the horizontal. If his
displacement is 100 m, then the coordinates
of the superhero are:
x = ( 100 m ) cos ( −30.0° ) = 86.6 m
y = ( 100 m ) sin ( −30.0° ) = −50.0 m
ANS. FIG. P3.52
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
120
P3.53
Vectors
(a)
Take the x axis along the tail section of the snake. The
displacement from tail to head is
( 240 m ) î + [(420 − 240) m ] cos(180° − 105°)î
− ( 180 m ) sin 75° ĵ = 287 mî − 174 mĵ
Its magnitude is
From v =
( 287 )2 + (174)2
m = 335 m.
distance
, the time for each child’s run is
Δt
distance 335 m ( h ) ( 1 km )( 3600 s )
=
= 101 s
v
(12 km )(1000 m )(1 h )
420 m ⋅ s
Olaf: Δt =
= 126 s
3.33 m
Inge: Δt =
Inge wins by 126 − 101 = 25.4 s .
(b)
Olaf must run the race in the same time:
v=
P3.54
d 420 m ⎛ 3600 s ⎞ ⎛ km ⎞
=
⎜
⎟⎜
⎟ = 15.0 km/h
Δt 101 s ⎝ 1 h ⎠ ⎝ 103 m ⎠
The position vector from the ground under the controller of the first
airplane is

r1 = (19.2 km)(cos 25°)î + (19.2 km)(sin 25°) ĵ + (0.8 km)k̂
(
)
= 17.4î + 8.11ĵ + 0.8k̂ km
The second is at

r2 = (17.6 km)(cos 20°)î + (17.6 km)(sin 20°) ĵ + (1.1 km)k̂
(
)
= 16.5î + 6.02 ĵ + 1.1k̂ km
Now the displacement from the first plane to the second is  
r2 − r1 = −0.863î − 2.09 ĵ + 0.3k̂ km
(
)
with magnitude ( 0.863)2 + ( 2.09)2 + ( 0.3)2
km = 2.29 km
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.55
(a)
121
The tensions Tx and Ty act as an equivalent tension T (see ANS.
FIG. P3.55) which supports the downward weight; thus, the
combination is equivalent to 0.150 N, upward. We know that Tx =
0.127 N, and the tensions are perpendicular to each other, so their
combined magnitude is
T = Tx2 + Ty2 = 0.150 N → Ty2 = ( 0.150 N ) − Tx2
2
Ty2 = ( 0.150 N ) − ( 0.127 N ) → Ty = 0.078 N
2
2
ANS. FIG. P3.55
(b)
(
)
From the figure, θ = tan −1 Ty Tx = 32.1°. The angle the x axis
makes with the horizontal axis is 90° − θ = 57.9° .
P3.56
(c)
From the figure, the angle the y axis makes with the horizontal
axis is θ = 32.1° .
(a)
Consider the rectangle in the figure to have height H and width

   

W. The vectors A and B are related by A + ab + bc = B, where

A = ( 10.0 m )( cos 50.0° ) î + ( 10.0 m )( sin 50.0° ) ĵ

A = 6.42 î + 7.66 ĵ m

B = ( 12.0 m )( cos 30.0° ) î + ( 12.0 m )( sin 30.0° ) ĵ

B = 10.4î + 6.00 ĵ m


ab = −Hĵ and bc = W î
(
)
(
)
ANS. FIG. P3.56
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
122
Vectors
Therefore,
   
B − A = ab + bc
( 3.96î − 1.66 ĵ) m = Wî − Hĵ → W = 3.96 m and H = 1.66 m
The perimeter measures 2(H + W) = 11.24 m.
(b)
The vector from the origin to the upper-right corner of the
rectangle (point d) is

B + Hĵ = 10.4 mî + ( 6.00 m + 1.66 m ) ĵ = 10.4 mî + 7.66 mĵ
(10.4 m )2 + (7.66 m )2 = 12.9 m
tan −1 ( 7.66/10.4 ) = 36.4° above + x axis ( first quadrant )
magnitude:
direction:
P3.57
(a)
Rx = 2.00 , Ry = 1.00 , Rz = 3.00
(b)

R = Rx2 + Ry2 + Rz2 = 4.00 + 1.00 + 9.00 = 14.0 = 3.74
(c)
⎛R ⎞
R
cos θ x = x ⇒ θ x = cos −1 ⎜ x ⎟ = 57.7° from + x
⎜⎝ R ⎟⎠
R
⎛ Ry ⎞
Ry
cos θ y =  ⇒ θ y = cos −1 ⎜  ⎟ = 74.5° from + y
⎜⎝ R ⎟⎠
R
⎛R ⎞
R
cos θ z = z ⇒ θ z = cos −1 ⎜ z ⎟ = 36.7° from + z
⎜⎝ R ⎟⎠
R
P3.58
Let A represent the distance from island 2 to

island 3. The displacement is A = A at 159°.
Represent the displacement from 3 to 1 as
 

B = B at 298°. We have 4.76 km at 37° + A + B = 0.
For the x components:
( 4.76 km ) cos 37° + A cos159°
+Bcos 298° = 0
ANS. FIG. P3.58
3.80 km − 0.934A + 0.470B = 0
B = −8.10 km + 1.99A
For the y components:
( 4.76 km ) sin 37° + A sin 159° + Bsin 298° = 0
2.86 km + 0.358A − 0.883B = 0
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
(a)
123
We solve by eliminating B by substitution:
2.86 km + 0.358A − 0.883 ( −8.10 km + 1.99A ) = 0
2.86 km + 0.358A + 7.15 km − 1.76A = 0
10.0 km = 1.40A
A = 7.17 km
(b)
P3.59
B = −8.10 km + 1.99 ( 7.17 km ) = 6.15 km
Let θ represent the angle between the




directions of A and B . Since A and B have
 
  
the same magnitudes, A , B , and R = A + B
form an isosceles triangle in which the angles

θ
θ
are 180° − θ , , and . The magnitude of R
2
2
⎛θ⎞
is then R = 2A cos ⎜ ⎟ . This can be seen from
ANS. FIG. P3.59
⎝ 2⎠
applying the law of cosines to the isosceles triangle and using the fact
that B = A. 
  

Again, A , −B , and D = A − B form an isosceles triangle with apex
angle θ. Applying the law of cosines and the identity ⎛θ ⎞
1 − cosθ = 2 sin 2 ⎜ ⎟
⎝ 2⎠

⎛θ⎞
gives the magnitude of D as D = 2A sin ⎜ ⎟ . ⎝ 2⎠
The problem requires that R = 100D. ⎛θ⎞
⎛θ⎞
Thus, 2A cos ⎜ ⎟ = 200A sin ⎜ ⎟ . This gives
⎝ 2⎠
⎝ 2⎠
⎛θ ⎞
tan ⎜ ⎟ = 0.010 and θ = 1.15°
⎝ 2⎠
P3.60
Let θ represent the angle between the directions




of A and B . Since A and B have the same
 
  
magnitudes, A , B , and R = A + B form an
isosceles triangle in which the angles are

θ
θ
180° − θ , , and . The magnitude of R is then
2
2
⎛θ⎞
R = 2A cos ⎜ ⎟ . This can be seen by applying the
⎝ 2⎠
ANS. FIG. P3.60
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124
Vectors
law of cosines to the isosceles triangle and using the fact that B = A.

  

Again, A , −B , and D = A − B form an isosceles triangle with apex
angle θ. Applying the law of cosines and the identity ⎛θ ⎞
1 − cosθ = 2 sin 2 ⎜ ⎟
⎝ 2⎠

⎛θ⎞
gives the magnitude of D as D = 2A sin ⎜ ⎟ . ⎝ 2⎠
The problem requires that R = nD or ⎛θ⎞
⎛θ⎞
⎛ 1⎞
cos ⎜ ⎟ = nsin ⎜ ⎟ giving θ = 2 tan −1 ⎜ ⎟ . ⎝ 2⎠
⎝ 2⎠
⎝ n⎠
P3.61

The larger R is to be compared to D, the smaller the angle between A

and B becomes. 
(a) We write B in terms of the sine and cosine of the angle θ , and
add the two vectors:
 
A + B = −60 cmĵ + ( 80 cm cosθ ) î + ( 80 cm sinθ ) ĵ
 
A + B = ( 80 cm cosθ ) î + ( 80 cm sinθ − 60 cm ) ĵ
(
)
Dropping units (cm), the magnitude is
 
2
2 1/2
A + B = ⎡⎣( 80 cosθ ) + ( 80 sinθ − 60 ) ⎤⎦
2
2
= ⎡⎣( 80 ) ( cos 2 θ + sin 2 θ ) − 2 ( 80 )( 60 ) sinθ + ( 60 ) ⎤⎦
1/2
 
1/2
2
2
A + B = ⎡⎣( 80 ) + ( 60 ) − 2 ( 80 )( 60 ) sinθ ⎤⎦
 
1/2
A + B = [ 10, 000 − ( 9600 ) sin θ ] cm
(b)
 
For θ = 270°, sinθ = −1, and A + B = 140 cm .
(c)
 
For θ = 90°, sinθ = 1, and A + B = 20.0 cm .
(d)

They do make sense. The maximum value is attained when A

and B are in the same direction, and it is 60 cm + 80 cm. The


minimum value is attained when A and B are in opposite
directions, and it is 80 cm – 60 cm.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.62
125
We perform the integration:

0.380 s 
0.380 s
Δr = ∫0
v dt = ∫0
1.2 î m/s − 9.8tĵ m/s 2 dt
= 1.2t î m/s
0.380 s
0
(
t
− ( 9.8 ĵ m/s )
2
2
)
2 0.380 s
0
⎛ ( 0.38 s )2 − 0 ⎞
= 1.2 î m/s ( 0.38 s − 0 ) − 9.8 ĵ m/s ⎜
⎟⎠
2
⎝
(
)
(
2
)
= 0.456î m − 0.708 ĵ m
P3.63
(
)
(a)

d r d 4î + 3 ĵ − 2tk̂
=
= −2 k̂ = − ( 2.00 m/s ) k̂
dt
dt
(b)
The position vector at t = 0 is 4î + 3 ĵ. At t = 1 s, the position is
4î + 3 ĵ − 2 k̂, and so on. The object is moving straight downward

dr
at 2 m/s, so
represents its velocity vector .
dt
P3.64
(a)
The very small differences
between the angles suggests
we may consider this region of
Earth to be small enough so
that we may consider it to be
flat (a plane); therefore, we
may consider the lines of
latitude and longitude to be
parallel and perpendicular, so
that we can use them as an xy
coordinate system. Values of
latitude, θ, increase as we
ANS. FIG. P3.64
travel north, so differences
between latitudes can give the y coordinate. Values of longitude,
φ, increase as we travel west, so differences between longitudes
can give the x coordinate. Therefore, our coordinate system will
have +y to the north and +x to the west.
Since we are near the equator, each line of latitude and longitude
may be considered to form a circle with a radius equal to the
radius of Earth, R = 6.36 × 106 m. Recall the length s of an arc of a
circle of radius R that subtends an angle (in radians) Δθ (or Δφ) is
given by s = RΔθ (or s = RΔφ ). We can use this equation to find
the components of the displacement from the starting point to the
tree—these are parallel to the x and y coordinates axes. Therefore,
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
126
Vectors
we can regard the origin to be the starting point and the
displacements as the x and y coordinates of the tree.
The angular difference Δφ for longitude values is (west being
positive)
Δφ = [75.64426° − 75.64238° ]
= ( 0.00188° )(π rad /180° )
= 3.28 × 10−5 rad
corresponding to the x coordinate (displacement west)
x = RΔφ = ( 6.36 × 106 m ) ( 3.28 × 10−5 rad ) = 209 m
The angular difference Δθ for latitude values is (north being
positive)
Δθ = [ 0.00162° − ( −0.00243° )]
= ( 0.00405° )(π rad /180° )
= 7.07 × 10−5 rad
corresponding to the y coordinate (displacement north)
y = RΔθ = ( 6.36 × 106 m ) ( 7.07 × 10−5 rad ) = 450 m
The distance to the tree is
d = x 2 + y 2 = ( 209 m ) + ( 450 m ) = 496 m
2
2
The direction to the tree is
⎛ y⎞
⎛ 450 m ⎞
tan −1 ⎜ ⎟ = tan −1 ⎜
= 65.1° = 65.1° N of W
⎝ x⎠
⎝ 209 m ⎟⎠
(b)
Refer to the arguments above. They are justified because the
distances involved are small relative to the radius of Earth.
P3.65
(a)

From the picture, R 1 = aî + bĵ. (b)
R 1 = a 2 + b 2 (c)


R 2 = R 1 + ck̂ = aî + bĵ + ck̂ ANS. FIG. P3.65
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
P3.66
Since 127
 
A + B = 6.00 ĵ, we have ( Ax + Bx ) î + ( Ay + By ) ĵ = 0î + 6.00 ĵ giving
Ax + Bx = 0 → Ax = −Bx . ANS. FIG. P3.66
Because the vectors have the same magnitude and x components of
equal magnitude but of opposite sign, the vectors are reflections of
each other in the y axis, as shown in the diagram. Therefore, the two
vectors have the same y components: Ay = By = (1/2)(6.00) = 3.00 

Defining θ as the angle between either A or B and the y axis, it is seen
that 3.00
= 0.600 → θ = 53.1°
A
B 5.00


The angle between A and B is then φ = 2θ = 106° . cos θ =
Ay
=
By
=
Challenge Problem
P3.67
(a)
(
)
 
You start at point A: r1 = rA = 30.0î − 20.0 ĵ m. The displacement to B is  
rB − rA = 60.0î + 80.0 ĵ − 30.0î + 20.0 ĵ = 30.0î + 100 ĵ
(
)
You cover half of this, 15.0î + 50.0 ĵ , to move to 
r2 = 30.0î − 20.0 ĵ + 15.0î + 50.0 ĵ = 45.0î + 30.0 ĵ
Now the displacement from your current position to C is  
rC − r2 = −10.0î − 10.0 ĵ − 45.0î − 30.0 ĵ = −55.0î − 40.0 ĵ
You cover one-third, moving to 1
 

r3 = r2 + Δr23 = 45.0î + 30.0 ĵ + −55.0î − 40.0 ĵ = 26.7 î + 16.7 ĵ
3
(
)
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128
Vectors
The displacement from where you are to D is  
rD − r3 = 40.0î − 30.0 ĵ − 26.7 î − 16.7 ĵ = 13.3î − 46.7 ĵ
You traverse one-quarter of it, moving to 1
  1  
r4 = r3 + ( rD − r3 ) = 26.7 î + 16.7 ĵ + 13.3î − 46.7 ĵ
4
4
= 30.0î + 5.00 ĵ
(
)
The displacement from your new location to E is  
rE − r4 = −70.0î + 60.0 ĵ − 30.0î − 5.00 ĵ = −100î + 55.0 ĵ of which you cover one-fifth the distance, −20.0î + 11.0 ĵ,
moving to 

r4 + Δr45 = 30.0î + 5.00 ĵ − 20.0î + 11.0 ĵ = 10.0î + 16.0 ĵ
The treasure is at ( 10.0 m, 16.0 m ) . (b)
Following the directions brings you to the average position of the
trees. The steps we took numerically in part (a) bring you to  
 1  
⎛ rA + rB ⎞
rA + ( rB − rA ) = ⎜
⎝ 2 ⎟⎠
2
then to  
( rA + rB )
2

 
  
rC − ( rA + rB ) / 2 rA + rB + rC
+
=
3
3
then to ( rA + rB + rC )
3

  
   
rD − ( rA + rB + rC ) / 3 rA + rB + rC + rD +
=
4
4
and last to ( rA + rB + rC + rD ) + rE − ( rA + rB + rC + rD ) / 4
4
5
     rA + rB + rC + rD + rE
=
5
This center of mass of the tree distribution is the same location
whatever order we take the trees in.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 3
129
ANSWERS TO EVEN-NUMBERED PROBLEMS
P3.2
(a) 2.31; (b) 1.15
P3.4
(a) (2.17, 1.25) m, (−1.90, 3.29) m; (b) 4.55m
P3.6
P3.8
(a) r, 180° – θ; (b) 180° + θ ; (c) –θ

B is 43 units in the negative y direction
P3.10
9.5 N, 57° above the x axis
P3.12
(a) See ANS. FIG. P3.12; (b) The sum of a set of vectors is not affected
by the order in which the vectors are added.
P3.14
310 km at 57° S of W
P3.16
Ax = 28.7 units, Ay = –20.1 units
P3.18
1.31 km north and 2.81 km east
P3.20
(a) 5.00 blocks at 53.1° N of E; (b) 13.00 blocks
P3.22
( 2.60î + 4.50 ĵ) m
P3.24
788 miles at 48.0° northeast of Dallas
P3.26
(a) See ANS. FIG. P3.24; (b) 5.00î + 4.00 ĵ, −1.00î + 8.00 ĵ; (c) 6.40 at 38.7°,
8.06 at 97.2°
P3.28
(a) Its component parallel to the surface is (1.50 m) cos 141° = −1.17 m,
or 1.17 m toward the top of the hill; (b) Its component perpendicular to
the surface is (1.50 m) sin 141° = 0.944 m, or 0.944 m away from the
snow.
P3.30
42.7 yards
P3.32
Cx = 7.30 cm; Cy = −7.20 cm
P3.34
59.2° with the x axis, 39.8° with the y axis, 67.4° with the z axis
P3.36
(a) 5.00î − 1.00 ĵ − 3.00 k̂, 5.92 m; (b) (4.00î − 11.0 ĵ + 15.0 k̂) m, 19.0) m
P3.38
(a) 49.5î + 27.1ĵ; (b) 56.4, 28.7°
P3.40
magnitude: 170.1 cm, direction: 57.2° above +x axis (first quadrant);
magnitude: 145.7 cm, direction: 58.6° above +x axis (first quadrant)
P3.42
(a) 3.12 î + 5.02 ĵ − 2.20 k̂ km; (b) 6.31 km
P3.44
Impossible because 12.4 m is greater than 5.00 m
(
)
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130
P3.46
Vectors
(a) ( 5 = 11 f ) î + ( 3 + 9 f ) ĵ meters; (b) ( 5 + 0 ) î + ( 3 + 0 ) ĵ meters; (c) This
is reasonable because it is the location of the starting point, 5î + 3 ĵ
meters. (d) 16î + 12 ĵ meters; (e) This is reasonable because we have
completed the trip, and this is the position vector of the endpoint.
P3.48
2.24 m, 26.6°
P3.50
390 mi/h at 7.37° N of E
P3.52
86.6 m, –50.0 m
P3.54
2.29 km
P3.56
(a) The perimeter measures 2(H + W) = 11.24 m; (b) magnitude: 12.9 m,
direction: 36.4° above +x axis (first quadrant)
P3.58
(a) 7.17 km; (b) 6.15 km
P3.60
⎛ 1⎞
θ = 2 tan −1 ⎜ ⎟
⎝ n⎠
P3.62
0.456î m − 0.708 ĵ m
P3.64
(a) 496 m, 65.1° N of W; (b) The arguments are justified because the
distances involved are small relative to the radius of the Earth.
P3.66
φ = 2θ = 106°
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.